# Ma1506 Leture Notes

Topics: Trigraph, Complex number, Linear algebra Pages: 17 (4824 words) Published: April 11, 2013
MA1506 LECTURE NOTES CHAPTER 1 DIFFERENTIAL EQUATIONS 1.1 Introduction

A diﬀerential equation is an equation that contains one or more derivatives of a diﬀerentiable function. [In this chapter we deal only with ordinary DEs, NOT partial DEs.] The order of a d.e. is the order of the equation’s highest order derivative; and a d.e. is linear if it can be put in the form

any (n)(x)+an−1y (n−1)(x)+· · ·+a1y (1)(x)+a0y(x) = F, 1

where ai, 0 ≤ i ≤ n, and F are all functions of x. For example, y = 5y and xy − sin x = 0 are ﬁrst order linear d.e.; (y )2 + (y )5 − y = ex is third order, nonlinear. We observe that in general, a d.e. has many solutions, e.g. y = sin x + c, c an arbitrary constant, is a solution of y = cos x. Such solutions containing arbitrary constants are called general solution of a given d.e.. Any solution obtained from the general solution by giving speciﬁc values to the arbitrary constants is called a particular solution of that d.e. e.g.

2

y = sin x + 1 is a particular solution of y = cos x. Basically, diﬀerential equations are solved using integration, and it is clear that there will be as many integrations as the order of the DE. Therefore, THE GENERAL SOLUTION OF AN nth-ORDER DE WILL HAVE n ARBITRARY CONSTANTS.

1.2

Separable equations

A ﬁrst order d.e. is separable if it can be written in the form M (x) − N (y)y = 0 or equivalently, M (x)dx = N (y)dy. When we write the 3

d.e. in this form, we say that we have separated the variables, because everything involving x is on one side, and everything involving y is on the other. We can solve such a d.e. by integrating w.r.t. x:

M (x)dx = N (y)dy + c. Example 1. Solve y = (1 + y 2)ex.

Solution. We separate the variables to obtain 1 e dx = dy. 1 + y2 x

4

Integrating w.r.t. x gives ex = tan−1 y + c, or tan−1 y = ex − c, or y = tan(ex − c). Example 2. Experiments show that a ra-

dioactive substance decomposes at a rate proportional to the amount present. Starting with 2 mg at certain time, say t = 0, what can be said about the amount available at a later time? Example 3. A copper ball is heated to 5

100◦C. At t = 0 it is placed in water which is maintained at 30◦C. At the end of 3 mins the temperature of the ball is reduced to 70◦C. Find the time at which the temperature of the ball is 31◦C. Physical information: Experiments show that the rate of change dT /dt of the temperature T of the ball w.r.t. time is proportional to the difference between T and the temp T0 of the surrounding medium. Also, heat ﬂows so rapidly in copper that at any time the temperature is practically the same at all points of the ball.

6

Example 4.

Suppose that a sky diver falls

from rest toward the earth and the parachute opens at an instant t = 0, when sky diver’s speed is v(0) = v0 = 10 m/s. Find the speed of the sky diver at any later time t. Physical assumptions and laws: weight of the man + equipment = 712N, air resistance = bv 2, where b = 30 kg/m. Using Newton’s second law we obtain m Thus dv = mg − bv 2. dt

dv b mg = − (v 2 − k 2), k 2 = dt m b 1 b dv = − dt v2 − k2 m 7

1  1 1  b   dv = − − dt. 2k v − k v + k m  

Integrating gives 2kb v−k =− n t + c1, v+k m or 2kb v−k −pt = ce , p = . v+k m (c = ±ec1 according as the ratio on the left is positive or negative). 1 + ce−pt Solving for v : v = k . (Note that 1 − ce−pt v → k as t → ∞). From v(0) = v0, c = v0 − k mg 712 2 . Now k = = , so k = 4.87 m/s, v0 + k b 30 v0 = 10 m/s, c = 0.345, p = 4.02. 1 + 0.345e−4.02t · · v(t) = 4.87 . 1 − 0.345e−4.02t ·

8

Example 5. A conference room with volume 2000m3 contains air with 0.002% CO. At time t = 0 the ventilation system starts blowing in air which contains 3% CO (by volume). If the ventilation system blows in (and extracts) air at a rate of 0.2m3/min, how long will it take for the air in the room to contain 0.015% CO? Solution. Let x m3 of CO be in the room at

time t. Then

dx dt

= in ﬂow − out ﬂow
x =...